19_ch 06 Mechanical Design budynas_SM_ch06

19_ch 06 Mechanical Design budynas_SM_ch06 - 6 ) + 4545 P...

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Chapter 6 165 6-27 S y = 800 MPa, S ut = 1000 MPa (a) From Fig. 6-20, for a notch radius of 3 mm and S ut = 1GPa, q . = 0 . 92 . K f = 1 + q ( K t 1) = 1 + 0 . 92(3 1) = 2 . 84 σ max =− K f 4 P π d 2 =− 2 . 84(4) P π (0 . 030) 2 =− 4018 P σ m =− σ a = 1 2 ( 4018 P ) =− 2009 P T = fP ± D + d 4 ² T max = 0 . 3 P ± 0 . 150 + 0 . 03 4 ² = 0 . 0135 P From Fig. 6-21, q s . = 0 . 95 . Also, K ts is given as 1.8. Thus, K fs = 1 + q s ( K ts 1) = 1 + 0 . 95(1 . 8 1) = 1 . 76 τ max = 16 K fs T π d 3 = 16(1 . 76)(0 . 0135 P ) π (0 . 03) 3 = 4482 P τ a = τ m = 1 2 (4482 P ) = 2241 P Eqs. (6-55) and (6-56): σ ± a = σ ± m = ³ ( σ a / 0 . 85) 2 + 3 τ 2 a ´ 1 / 2 = ³ ( 2009 P / 0 . 85) 2 + 3(2241 P ) 2 ´ 1 / 2 = 4545 P S ± e = 0 . 5(1000) = 500 MPa k a = 4 . 51(1000) 0 . 265 = 0 . 723 k b = ± 30 7 . 62 ² 0 . 107 = 0 . 864 S e = 0 . 723(0 . 864)(500) = 312 . 3MPa Modified Goodman: σ ± a S e + σ ± m S ut = 1 n 4545 P 312 . 3(10
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Unformatted text preview: 6 ) + 4545 P 1000(10 6 ) = 1 3 P = 17 . 5(10 3 ) N = 16 . 1 kN Ans . Yield (conservative): n y = S y a + m n y = 800(10 6 ) 2(4545)(17 . 5)(10 3 ) = 5 . 03 Ans . (actual): max = ( 2 max + 3 2 max ) 1 / 2 = ( 4018 P ) 2 + 3(4482 P ) 2 1 / 2 = 8741 P n y = S y max = 800(10 6 ) 8741(17 . 5)10 3 = 5 . 22...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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